(2) Calculate the spring constant, k, for the spring. State an appropriate unit.

Transcription

1 Q1. A manufacturer of springs tests the properties of a spring by measuring the load applied each time the extension is increased. The graph of load against extension is shown below. (a) State Hooke s law. () (b) Calculate the spring constant, k, for the spring. State an appropriate unit. spring constant... unit... (3) (c) Use the graph to find the work done in extending the spring up to point B. work done... J (3) Page 1 of 1

2 (d) Beyond point A the spring undergoes plastic deformation. Explain the meaning of the term plastic deformation. (e) (f) When the spring reaches an extension of m, the load on it is gradually reduced to zero. On the graph above sketch how the extension of the spring will vary with load as the load is reduced to zero. Without further calculation, compare the total work done by the spring when the load is removed with the work that was done by the load in producing the extension of m. (1) () (1) (Total 1 marks) Page of 1

3 Q. The diagram below shows a tower crane that has two identical steel cables. The length of each steel cable is 35 m from the jib to the hook. (a) Each cable has a mass of 4.8 kg per metre. Calculate the weight of a 35 m length of one cable. weight =... N () (b) The cables would break if the crane attempted to lift a load of N or more. Calculate the breaking stress of one cable. cross-sectional area of each cable = m breaking stress =... Pa () Page 3 of 1

4 (c) When the crane supports a load each cable experiences a stress of 400 MPa. Each cable obeys Hooke s law. Ignore the weight of the cables. Young modulus of steel = Pa (i) Calculate the weight of the load. weight =... N () (ii) The unstretched length of each cable is 35 m. Calculate the extension of each cable when supporting the load. extension =... m (3) (iii) Calculate the combined stiffness constant, k, for the two cables. stiffness constant =... Nm 1 () (iv) Calculate the total energy stored in both stretched cables. energy stored =... J () (Total 13 marks) Page 4 of 1

5 Q3. (a) Describe an experiment to accurately determine the spring constant k of a spring that is thought to reach its limit of proportionality when the load is about 0 N. Include details of the necessary measurements and calculations and describe how you would reduce uncertainty in your measurements. A space is provided for a labelled diagram should you wish to include one. The quality of your written communication will be assessed in this question. (6) Page 5 of 1

6 (b) Two identical springs, each having a spring constant of 85 Nm 1, are shown arranged in parallel and series in the figure below. A load of 15 N is attached to each arrangement. (i) Calculate the extension for the parallel arrangement when the load is midway between the lower ends of the springs. answer =... m () (ii) Calculate the extension for the series arrangement. answer =... m () (iii) Calculate the energy stored in the parallel arrangement. answer =... J () Page 6 of 1

7 (iv) Without further calculation, discuss whether the energy stored in the series arrangement is less, or greater, or the same as in the parallel arrangement (3) (Total 15 marks) Q4. The figure below shows a stress-strain graph for a copper wire. (a) Define tensile strain. (1) (b) State the breaking stress of this copper wire. (c) answer =... Pa Mark on the figure above a point on the line where you consider plastic deformation may start. Label this point A. (1) (1) Page 7 of 1

8 (d) Use the graph to calculate the Young modulus of copper. State an appropriate unit for your answer. answer =... (3) (e) The area under the line in a stress-strain graph represents the work done per unit volume to stretch the wire. (i) Use the graph to find the work done per unit volume in stretching the wire to a strain of answer =...J m 3 () (ii) Calculate the work done to stretch a kg sample of this wire to a strain of The density of copper = 8960 kg m 3. answer =...J () (f) A certain material has a Young modulus greater than copper and undergoes brittle fracture at a stress of 176 MPa. On the figure above draw a line showing the possible variation of stress with strain for this material. () (Total 1 marks) Page 8 of 1

9 Q5. The table below shows the results of an experiment where a force was applied to a sample of metal. (a) On the axes below, plot a graph of stress against strain using the data in the table. Strain / Stress /10 8 Pa (3) (b) Use your graph to find the Young modulus of the metal. answer =... Pa () Page 9 of 1

10 (c) A 3.0 m length of steel rod is going to be used in the construction of a bridge. The tension in the rod will be 10 kn and the rod must extend by no more than 1.0mm. Calculate the minimum cross-sectional area required for the rod. Young modulus of steel = Pa answer =... m (3) (Total 8 marks) Q6. (a) The speed of light is given by c = f λ State how each of these quantities will change, if at all, when light travels from air to glass. c... f... λ... (3) Figure 1 shows a side view of a step index optical fibre. Figure 1 (b) Ray A enters the end of the fibre and then undergoes total internal reflection. On Figure 1 complete the path of this ray along the fibre. () Page 10 of 1

11 (c) (i) The speed of light in the core is ms 1. Show that the refractive index of the core is () (ii) Show that the critical angle at the boundary between the core and the cladding is about 80. refractive index of the cladding = 1.45 () (d) Ray B enters the end of the fibre and refracts along the core-cladding boundary. Calculate the angle of incidence, θ, of this ray at the point of entry to the fibre. answer =... degrees (3) Page 11 of 1

12 (e) Figure shows a pulse of monochromatic light (labelled X) that is transmitted a significant distance along the fibre. The shape of the pulse after travelling along the fibre is labelled Y. Explain why the pulse at Y has a lower amplitude and is longer than it is at X. Figure () (Total 14 marks) Page 1 of 1

13 M1. (a) Force proportional to extension up to the limit of proportionality (accept elastic limit) dependent upon award of first mark Symbols must be defined Accept word equation allow F=kΔL (or F ΔL) up to the limit of proportionality for the second mark only allow stress strain up to the limit of proportionality for the second mark only (b) Gradient clearly attempted / use of k=f / ΔL k = 30 / 0.06 = 1154 or 31 / 0.07 = 1148 correct values used to calculate gradient with appropriate sf answer given (1100 or 100) 1100 or 100 with no other working gets 1 out of OR 1154 ± 6 seen Do not allow 3/0.080 or 33/0.090 (point A) for second mark. AND load used >= 15 (= 1100 or 100 (sf) ) 3 / 0.08 is outside tolerance. 3/0.077 is just inside. Nm 1 / N / m (newtons per metre) (not n / m, n / M, N / M) 3 (c) any area calculated or link energy with area / use of 1 / FΔL (or Nm for little squares) 35 whole squares, 16 part gives 43 ± 1.0 OR equivalent correct method to find whole area (d) 0.05 Nm per (1cm) square candidates number of squares and correctly evaluated OR (= 1.075) = 1.1 (J) (1.05 to 1.10 if not rounded) permanent deformation / permanent extension Allow: doesn t return to original length ; correct reference to yield e.g. allow extension beyond the yield point do not accept: does not obey Hooke s law or ceases to obey Hooke s law, 3 1 (e) any line from B to a point on the x axis from to 0.00 straight line from B to x axis (and no further) that reaches x axis for 0.010<=ΔL<= (f) work done by spring < work done by the load Accept less work or it is less (we assume they are referring to the work done by spring) 1 [1] Page 13 of 1

14 M. (a) (W = mg) = =1600 (1648 N) Allow g=10 : 1680 (1700 N) g = N max 1 for doubling or halving. Max 1 for use of grammes (b) (stress = tension / area) For first mark, forgive absence of or incorrect doubling / halving. = (0.5 ) / OR = / ( ) = (1.1 GPa) Forgive incorrect prefix if correct answer seen. (c) (i) (weight = stress area) max 1 mark for incorrect power of ten in first marking point = 400 (10 6 ) (= N) max 1 mark for doubling or halving both stress and area ( = ) ( N) Forgive incorrect prefix if correct answer seen.look out for YM 400k Pa which gives correct answer but scores zero. (ii) OR correct substitution into a correct equation (forgive incorrect doubling or halving for this mark only OR alternative method: strain = stress / E then ΔL = L strain If answer to 4ci is used, it must be halved, unless area is doubled, for this mark Any incorrect doubling or halving is max 1 mark. Allow Page 14 of 1

15 (iii) OR correct substitution into F=kΔL answer 4c(ii) ) ecf ci and cii (answer 4c(i) Allow halving extension for force on one cable = 7.4(4) 10 6 (Nm 1 ) Correct answer gains both marks (iv) Correct answer gains both marks = ½ OR ½ 7.4(4) 10 6 ( ) ecf ci, cii, ciii = 1.6(5) 10 4 (J) Forgive incorrect prefix if correct answer seen. Doubling the force gets zero. [13] M3. (a) The candidate s writing should be legible and the spelling, punctuation and grammar should be sufficiently accurate for the meaning to be clear. The candidate s answer will be assessed holistically. The answer will be assigned to one of three levels according to the following criteria. High Level (Good to excellent): 5 or 6 marks The information conveyed by the answer is clearly organised, logical and coherent, using appropriate specialist vocabulary correctly. The form and style of writing is appropriate to answer the question. Candidate must suggest drawing a graph of F vs ΔL (or vice versa) AND that k is in some way linked to the gradient AND use of a suitable named instrument to measure or determine extension AND 1 further means of reducing uncertainty: repeats / minimum 8 different readings / use of vernier scale / check values of mass with balance / parallax elimination with set square, pointer in contact with scale, mirror. Page 15 of 1

16 For 6 marks: must also give suitable range at least up to 10N but not beyond 0N (accept up to 0N / not beyond 0N ) AND minimum 8 different readings OR parallax elimination must be included AND repeats must be included AND correctly explains how k is obtained from their graph. Intermediate Level (Modest to adequate): 3 or 4 marks The information conveyed by the answer may be less well organised and not fully coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used incorrectly. The form and style of writing is less appropriate. Candidate must suggest: to measure / determine extension OR initial and final length AND to use F = k ΔL or k = F / ΔL OR drawing a graph of F vs ΔL (or vice versa) AND use of suitable instrument to measure extension OR 1 means of reducing uncertainty: repeats / use of vernier scale / check values of mass with balance / parallax elimination with set square, pointer in contact with scale, mirror / minimum 8 different readings / graphical approach For 4 marks, uncertainty comment AND instrument required Low Level (Poor to limited): 1 or marks The information conveyed by the answer is poorly organised and may not be relevant or coherent. There is little correct use of specialist vocabulary. The form and style of writing may be only partly appropriate. Any relevant statement from the marking points above For marks: must mention minimum two points including: to measure / determine extension OR initial and final length 6 (b) (i) (k = 85 = 170 (N m 1 ) ) (ΔL = F / k =) 15 / 170 ( or 7.5 / 85 ) = (m) (0.088) (ii) (k = ½ 85 = 4.5 ) (ΔL = F / k = ) 15/4.5 ( or x 15/85) = 0.35 (m) (0.359) Page 16 of 1

17 (iii) (W = ½ FΔL or ½ k ΔL ) = ½ ( or x ½ ) ecf 5bi = 0.66 (J) (0.6615) ecf 5bi (iv) (series) greater ecf for answer less or same where candidates incorrect answers to bi and bii support this. extension is more (in series) and the force is the same (in both situations) AND quotes Energy stored = ( ½ )Fs or ½ FΔL OR energy proportional to extension 3 [15] M4. (a) extension divided by its original length do not allow symbols unless defined 1 (b) (Pa) 1 (c) point on line marked A between a strain of and (d) clear evidence of gradient calculation for straight section eg 1.18 (1.) 10 8 / = 10 GPa and stress used > Pa allow range GPa Pa or Nm or N/m 3 (e) (i) clear attempt to calculate correct area (evidence on graph is sufficient) (3 whole squares + 1 part/ = 38 squares) ( = ) (J m 3 ) allow range to (ii) V = m/ρ or 0.015/8960 or (m 3 ) = 0.64 (0.636 J) ecf from ei Page 17 of 1

18 (f) straight line passing through origin (small curvature to the right only above 160 MPa is acceptable) end at 176 MPa (allow 174 to 178) straight section to the left of the line for copper (steeper gradient) [1] M5. (a) Suitable scale on both axes (eg not going up in 3s) and > ½ space used points correct (within half a small square) line is straight up to at least stress = and curve is smooth beyond straight section 3 (b) understanding that E = gradient (= Δy/Δx) allow y/x if line passes through origin = (Pa) (allow 0.90 to 1.1) ecf from their line in (a) if answer outside this range and uses a y value when values used from table; two marks can be scored only if candidates line passes through them one mark only can be scored if these points are not on their line Page 18 of 1

19 (c) correct rearrangement of symbols or numbers ignoring incorrect powers of ten, eg A = correct substitution in any correct form of the equation, eg = allow incorrect powers of ten for this mark = (1.5789) (m ) 3 [8] M6. (a) decrease constant decrease 3 (b) straight ray (ignore arrow) reflecting to the right reflected angle = incident angle (accept correct angle labels if reflected angle is outside tolerance) (c) (i) (n = ) use of 3 ( 10 8 ) = = 1.47 (1.4706) (must see 3 sf or more) (ii) sin θ c = or correct substitution in un-rearranged formula θ c = 80.4 (80.401) (80.3 to 80.54) ( 80 ) must see 3 sf or more Page 19 of 1

20 (d) angle of refraction = = 9.6 sinθ = 147(06) sin 9.6 = 0.5 ecf from first mark θ = 14 (= ) ecf from first mark range 13 to 15 due to use of rounded values 3 (e) (reduced amplitude) due to absorption/energy loss (within the fibre)/attenuation/scattering (by the medium) /loss from fibre (pulse broadening caused by) multi-path (modal) dispersion /different rays/modes propagating at different angles/non axial rays take longer time to travel same distance along fibre as axial rays [14] Page 0 of 1

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